Answer
$\frac{x+6}{(x^2+4)(x-2)}=\frac{1}{x-2}+\frac{-x-1}{x^2+4}$
Work Step by Step
$\frac{x+6}{x^3-2x^2+4x-8}$, factorising $x^3-2x^2+4x-8=x^2(x-2)+4(x-2)=(x^2+4)(x-2)$,
thus,
$\frac{x+6}{(x^2+4)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+4}$,
$A(x^2+4)+(Bx+C)(x-2)$,
$Ax^2+4A+Bx^2-2Bx+Cx-2C$,
$(A+B)x^2+(-2B+C)x+(4A-2C)=x+6$,
$\begin{cases}
A+B=0\\
-2B+C=1\\
4A-2C=6
\end{cases}$,
Adding Equation 2 and Equation 3 after simplifying by dividing both sides of the Equation by 2.
$\begin{cases}
-2B+C=1\\
2A-C=3\\
-- -- -- --\\
2A-2B=4
\end{cases}$
Adding Equation 1 and the new addition result after simplifying by dividing both sides of the Equation by 2.
$\begin{cases}
A+B=0\\
A-B=2\\
-- -- -- --\\
2A=2
\end{cases}$
thus, $A=1$, substituting back into the Equation, $1+B=0, B=-1$.
-substituting back into the Equation, $2+C=1, C=-1$.
Therefore,
$\frac{x+6}{(x^2+4)(x-2)}=\frac{1}{x-2}+\frac{-x-1}{x^2+4}$
